考虑层序遍历
如果当前节点的左右子树任一为空,判断同层的下一个节点的左右子树是否都为空即可
import java.util.*;
/*
* public class TreeNode {
* int val = 0;
* TreeNode left = null;
* TreeNode right = null;
* public TreeNode(int val) {
* this.val = val;
* }
* }
*/
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
*
* @param root TreeNode类
* @return bool布尔型
*/
int num = -1;
public boolean isCompleteTree (TreeNode root) {
if (root == null) {
return true;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
while (size-- > 0) {
TreeNode node = queue.poll();
if (node.left == null && node.right != null) {
return false;
}
if (num > -1 && (node.left != null || node.right != null)) {
return false;
}
if (node.left == null || node.right == null) {
num = 0;
}
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
}
return true;
}
}

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